3. The attempt at a solution
If the function were, for instance, [tex]g(x)=|x+1|-|x-1|[/tex], the solution wouldn't be a problem, because the two important points (x=-1 and x=1) can be recognized immediately, which implies analysing the three intervals ([tex]<-\infty,-1>[/tex] , [tex][-1,1>[/tex] , [tex][1,+\infty>[/tex]), and therefore the function g(x) can be seen as a compound of three different "sub-functions" on those intervals, ie:

The same should be done for [tex]f(x)=|x+|x+|x-1|||[/tex]. But how? Where to start? If starting from the "inside", there would be, at the first step, two cases: [tex]x-1\geq 0[/tex] or [tex]x-1<0[/tex], which would lead to more sub-cases, so I'm not sure if this is the right approach to arrive at the graph of f(x).

First find the key points where the function changes. There are only 2 points where it changes. For example, plug in 10 for x.
[tex]f(x)=|10+|10+|10-1|||[/tex]
f(x) = 29
the key points should not be that hard to find.